Let $k$ be an algebraicly closed field. Show that the set $C = \{ (t-1, t^2, t^3-t) | t \in k \} \subset k^3$ is an irreducible algebraic set.

I'm pretty sure I know how to do this: Find polynomials $f_1, \ldots, f_n$ such that $C = V((f_1, \ldots, f_n)$ and then somehow use the fact that those polynomials will probably be irreducible over $k[X, Y, Z]$. Am I correct in this assumption?

I have trouble finding such polynomials. Am I required to just "eyeball" some polynomials satisfying this or is there a different approach to the problem?


  • $\begingroup$ Are you sure the third coordinate isn't $t^3 - t^2$? Because, then $xy = z$ would be one equation. Anyway, one way to do this is to use Gröbner bases. $\endgroup$ Nov 1, 2016 at 15:40
  • $\begingroup$ Thats what it says, yeah. $xy-z$ was my first guess aswell, but that sadly doesn't work out. And I'm certain we do not have to use Gröbner bases. $\endgroup$ Nov 1, 2016 at 15:42
  • $\begingroup$ Okay, if you don't want to use Gröbner bases or resultants, then eyeballing it is. One easy one is $y=(x+1)^2$ and a slightly harder to see relation is $z = (x+1)^3 - x - 1$. $\endgroup$ Nov 1, 2016 at 15:51
  • $\begingroup$ Alright, so if I have $C = V((x+1)^2 - y, (x+1)^3 - x - z - 1))$, it is enough to say that this set is irreducible since the polynomials are irreducible, so $I(C)$ is a prime ideal? Or do I have to make additional computations first $\endgroup$ Nov 1, 2016 at 16:15

1 Answer 1


Shifting by $1$, we see that the same set can be described by $C'=\big(t, (t+1)^2, (t+1)^3-t-1)\big)$ .

Now consider the morphism $\phi\colon \mathbf{A}^1\to \mathbf{A}^2$ defined by $\phi(t)=( (t+1)^2, (t+1)^3-t-1)) $. Now it is clear that $C'$ is the graph of the morphism $\phi$ and hence is a closed irreducible subvariety of $ \mathbf{A}^3$.

  • $\begingroup$ Thanks for your answer, but I would like to do it the way I described, by using $I$ and $V$ and things like the radical. We have not even named varieties yet in the lecture $\endgroup$ Nov 1, 2016 at 16:15
  • $\begingroup$ In that case the comment by SpamIam has all that you need. $\endgroup$ Nov 1, 2016 at 16:21

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